Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int NMAX = 1e4;
const int LGMAX = 30;
vector<pair<int, int> > pref_op;
vector<pair<int, int> > suff_op;
int n;
int x[NMAX + 5];
int y[NMAX + 5];
int bx[LGMAX + 5];
int by[LGMAX + 5];
void do_op(int x[], int i, int j, vector<pair<int, int> > &op) {
  x[i] ^= x[j];
  op.push_back({i, j});
}
void sw(int x[], int i, int j, vector<pair<int, int> > &op) {
  if (i == j) {
    return;
  }
  do_op(x, i, j, op);
  do_op(x, j, i, op);
  do_op(x, i, j, op);
}
int do_gauss(int a[], int b[], vector<pair<int, int> > &op) {
  int l = 0;
  for (int i = 0; i <= LGMAX; i++) {
    for (int j = l + 1; j <= n; j++) {
      if ((a[j] >> i) & 1) {
        sw(a, ++l, j, op);
        b[l] = i;
        break;
      }
    }
    if (b[l] != i) {
      continue;
    }
    for (int j = 1; j <= n; j++) {
      if (j != l && (((a[j] & a[l]) >> i) & 1)) {
        do_op(a, j, l, op);
      }
    }
  }
  return l;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &y[i]);
  }
  int n1 = do_gauss(x, bx, pref_op);
  int n2 = do_gauss(y, by, suff_op);
  if (n1 < n2) {
    printf("-1\n");
    return 0;
  }
  int l = 0;
  for (int i = 1; i <= n2; i++) {
    if (bx[i] != by[i]) {
      do_op(x, i, i, pref_op);
    }
    for (int j = i + 1; j <= n1; j++) {
      if (((x[i] ^ y[i]) >> bx[j]) & 1) {
        do_op(x, i, j, pref_op);
      }
    }
  }
  for (int i = 1; i <= n1; i++) {
    if (y[i] == 0) {
      do_op(x, i, i, pref_op);
    }
    if (x[i] != y[i]) {
      printf("-1\n");
      return 0;
    }
  }
  while (suff_op.size()) {
    pref_op.push_back(suff_op.back());
    suff_op.pop_back();
  }
  printf("%d\n", (int)pref_op.size());
  for (auto it : pref_op) {
    printf("%d %d\n", it.first, it.second);
  }
  return 0;
}